Question

If a scuba diver releases a 16−mL air bubble below the surface where the pressure is 3.3 atm, what is the volume (in mL) of the bubble when it rises to the surface and the pressure is 1.0 atm?

Answer 1

$V_1 = 16 \;mL \\ P_1 = 3.3 \;atm \\ P_2 = 1.0 \;atm$

From Boyle’s Law

$P_1V_1=P_2V_2 \\ 3.3 \times 16 = 1 \times V_2 \\ V_2 = 52.8 \;mL$

Answer: 52.8 mL